Proof Let E be the least nonstationary subset of $\kappa $ in the canonical well-ordering of $\mathrm {{HOD}}$ that is $\Sigma _n(A)$ -stationary in $\kappa $ . Since the collection of all initial segments of the canonical well-ordering of $\mathrm {{HOD}}$ is definable by a $\Sigma _2$ -formula without parameters, and the collection $C^{(n)}$ of all ordinals $\lambda $ with $V_\lambda \prec _{\Sigma _n}V$ is definable by a $\Pi _n$ -formula without parameters (see [Reference Bagaria1, Section 1]), we know that the set $\{E\}$ is definable by a $\Sigma _{n+1}$ -formula with parameters in A. Let C denote the least closed unbounded subset of $\kappa $ in the canonical well-ordering of $\mathrm {{HOD}}$ that is disjoint from E. But then we know that the set $\{C\}$ is also definable by a $\Sigma _{n+1}$ -formula with parameters in A, and therefore, C witnesses that E is not $\Sigma _{n+1}(A)$ -stationary in $\kappa $ .

Proof By our assumptions, we can find a generic extension M of $\mathrm {{V}}$ such that $\mathcal {C}\in M\subsetneq \mathrm {{V}}[G]$ and $\mathrm {{V}}[G]$ is a nontrivial $\mathrm {{Add}}({\kappa },{\theta })$ -generic extension of M. Let $H\in \mathrm {{V}}[G]$ be $\mathrm {{Add}}({\kappa },{1})$ -generic over M, let X denote the subset of $\kappa $ corresponding to H, and set $E=X\setminus \mathrm {{Lim}}(X)$ . Genericity then ensures that E intersects every unbounded subset of $\kappa $ in M. In particular, we know that $C\cap E\neq \emptyset $ holds for every $C\in \mathcal {C}$ . Finally, since E is disjoint from $\mathrm {{Lim}}(X)$ , we know that E is a nonstationary subset of $\kappa $ in $\mathrm {{V}}[G]$ .

Proof of Theorem 2.1

Assume that the $\mathrm {{GCH}}$ holds. Given an uncountable regular cardinal $\kappa $ and a cardinal $\theta $ satisfying $\theta =\theta ^\kappa $ , we define $\mathbb {P}_{\kappa ,\theta }$ to be the two-step iteration $\mathrm {{Add}}({\kappa },{\theta })*\dot {{\mathbb {Q}}}$ , where $\dot {{\mathbb {Q}}}$ is the canonical $\mathrm {{Add}}({\kappa },{\theta })$ -name for the ${<}\theta ^+$ -closed partial order that codes ${\mathcal {P}}({\kappa })$ into the $\mathrm {{GCH}}$ -pattern above $\theta ^+$ (see, for example, [Reference Fuchs, Hamkins and Reitz7]). Let $G*H$ be ( $\mathrm {{Add}}({\kappa },{\theta })*\dot {{\mathbb {Q}}}$ )-generic over V. Fix a natural number $n>0$ and a set $A\in V[G,H]$ of cardinality less than $\theta $ with the property that, in $V[G,H]$ , the set $\{A\}$ is definable by a $\Sigma _2$ -formula with parameters in A. Since ${\mathcal {P}}({{\mathcal {P}}({\kappa })})^{V[G]}={\mathcal {P}}({{\mathcal {P}}({\kappa })})^{V[G,H]}$ and, in $V[G,H]$ , there are less than $\theta $ -many closed unbounded subsets C of $\kappa $ with the property that the set $\{C\}$ is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ , an application of Proposition 2.4 shows that, in $V[G,H]$ , there is a nonstationary subset of $\kappa $ that is $\Sigma _n(A)$ -stationary in $\kappa $ . But our setup ensures that ${\mathcal {P}}({\kappa })^{V[G]}={\mathcal {P}}({\kappa })^{V[G,H]}\subseteq \mathrm {{HOD}}^{V[G,H]}$ , and hence, Proposition 2.3 allows us to conclude that, in $V[G,H]$ , there is a subset of $\kappa $ that is $\Sigma _n(A)$ -stationary in $\kappa $ and not $\Sigma _{n+1}(A)$ -stationary in $\kappa $ .

Proof Let $\langle {C_\alpha }~\vert ~{\alpha <{\mathrm {{cof}}(\kappa )}}\rangle $ be an enumeration of all closed unbounded subsets C of $\kappa $ with the property that the set $\{C\}$ is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ . In addition, define C to be a diagonal intersection of these sets with respect to some strictly increasing continuous sequence of order type ${\mathrm {{cof}}(\kappa )}$ in $\kappa $ . Define $S=C\setminus \mathrm {{Lim}}(C)$ . Then S is an unbounded and nonstationary subset of $\kappa $ . Moreover, we have $C\cap S\neq \emptyset $ whenever C is a closed unbounded subset of $\kappa $ with the property that the set $\{C\}$ is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ .

Proof Since the set $[\kappa ]^\omega $ contains an almost disjoint family of cardinality $\kappa ^\omega $ that consists of unbounded subsets of $\kappa $ , we can find an element $b\in [\kappa ]^\omega $ that is unbounded in $\kappa $ and has the property that no infinite subset of b is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ . Then, $\kappa \setminus b$ is $\Sigma _n(A)$ -stationary in $\kappa $ and the complement of this set in $\kappa $ is unbounded in $\kappa $ .

Proof Since $A\cap \kappa $ is unbounded in $\kappa $ , we know that there are ${\mathrm {{cof}}(\kappa )}$ -many $\Sigma _n(A)$ -definable closed unbounded subsets of $\kappa $ . Let $\langle {C_\alpha }~\vert ~{\alpha <{\mathrm {{cof}}(\kappa )}}\rangle $ be an enumeration of all closed unbounded subsets C of $\kappa $ with the property that the set $\{C\}$ is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ . Since the assumption that $A\cap \kappa $ is unbounded in $\kappa $ implies that $C_\alpha \cap S$ is unbounded in $\kappa $ for all $\alpha <{\mathrm {{cof}}(\kappa )}$ , we can find a strictly increasing sequence $\langle {\sigma _\alpha }~\vert ~{\alpha <{\mathrm {{cof}}(\kappa )}}\rangle $ of elements of S with the property that $\sigma _{{\prec }{\alpha _0},{\alpha _1}{\succ }}\in C_{\alpha _1}$ for all $\alpha _0,\alpha _1<{\mathrm {{cof}}(\kappa )}$ , where ${{\prec }{\cdot },{\cdot }{\succ }}:{\mathrm{Ord}\times \mathrm{Ord}}\longrightarrow {\mathrm{Ord}}$ denotes the Gödel pairing function. If we now pick a partition $\langle {S_\alpha }~\vert ~{\alpha <{\mathrm {{cof}}(\kappa )}}\rangle $ of S with the property that $\{{\sigma _{\alpha ,\beta }}~\vert ~{\beta <{\mathrm {{cof}}(\kappa )}}\}\subseteq S_\alpha $ holds for all $\alpha <{\mathrm {{cof}}(\kappa )}$ , then each $S_\alpha $ is $\Sigma _n(A)$ -stationary in $\kappa $ .

Proof First, assume that (1) holds and (2) fails. Then, there is a singular cardinal $\kappa $ of countable cofinality and a subset A of $[\kappa ]^\omega $ such that A consists of cofinal subsets of $\kappa $ , $\vert {A}\vert \leq \theta $ , and for every subset E of $\kappa $ , there is $a\in A$ such that either $a\cap E$ or $a\setminus E$ is finite. Let $\langle {\kappa _n}~\vert ~{n<\omega }\rangle $ be a strictly increasing sequence that is cofinal in $\kappa $ with $\kappa _0=0$ . Given $a\in A$ , define

$$ \begin{align*}b_a ~ = ~ \{{n<\omega}~\vert~{a\cap[\kappa_n,\kappa_{n+1})\neq\emptyset}\} ~ \in ~ [\omega]^\omega.\end{align*} $$

Since $\vert {A}\vert <\mathfrak {r}$ , we can now find $c\in [\omega ]^\omega $ with the property that for all $a\in A$ , both $b_a\setminus c$ and $b_a\cap c$ are infinite. Set

$$ \begin{align*}E ~ = ~ \bigcup_{n\in c}[\kappa_n,\kappa_{n+1}).\end{align*} $$

By our assumptions, there is $a\in A$ with the property that either $a\setminus E$ or $a\cap E$ is finite. But this implies that either $b_a\setminus c$ or $b_a\cap c$ is finite, a contradiction.

Now, assume that (3) holds and (1) fails. Then, there is a singular cardinal $\kappa $ of countable cofinality with the property that for every subset A of $[\kappa ]^\omega $ that consists of cofinal subsets of $\kappa $ and has cardinality at most $\theta $ , there exists a subset E of $\kappa $ such that for every $a\in A$ , both $a\cap E$ and $a\setminus E$ are infinite. Fix $A\subseteq [\omega ]^\omega $ of cardinality $\mathfrak {r}$ such that for every $b\in [\omega ]^\omega $ , there exists $a\in A$ with the property that either $a\setminus b$ or $a\cap b$ is finite. Pick a s strictly increasing sequence $\langle {\kappa _n}~\vert ~{n<\omega }\rangle $ that is cofinal in $\kappa $ and, given $a\in A$ , define $b_a=\{{\kappa _n}~\vert ~{n\in a}\}\in [\kappa ]^\omega $ . Since $b_a$ is cofinal in $\kappa $ for all $a\in A$ , we can now find a subset E of $\kappa $ such that for every $a\in A$ , both $b_a\cap E$ and $b_a\setminus E$ are infinite. Set $c=\{{n<\omega }~\vert ~{\kappa _n\in E}\}\in [\omega ]^\omega $ . If $a\in A$ , then both $a\setminus c$ and $a\cap c$ are infinite, contradicting our assumptions on A.

Proof of Proposition 2.8

Let $\kappa $ be a singular cardinal of countable cofinality, let A be a set of cardinality less than $\mathfrak {r}$ , and let $n>0$ be a natural number. Then, there exists a subset $A'$ of $[\kappa ]^\omega $ of cardinality less than $\mathfrak {r}$ that consists of cofinal sequences and has the property that for every closed unbounded subset C of $\kappa $ such that the set $\{C\}$ is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ , there exists $a\in A'$ with $a\subseteq C$ . Using Lemma 2.9, we can now find a subset E of $\kappa $ such that for every $a\in A$ , both $a\setminus E$ and $a\cap E$ are infinite. Then, both E and $\kappa \setminus E$ are $\Sigma _n(A)$ -stationary in $\kappa $ .

Proof First, we assume that there is a cofinal function ${c}:{\omega }\longrightarrow {\kappa }$ that is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ . Define $C_0=\{{c(n)+\omega }~\vert ~{n<\omega }\}$ and $C_1=\{{c(n)+\omega +1}~\vert ~{n<\omega }\}$ . Then, $C_0$ and $C_1$ are disjoint closed unbounded in $\kappa $ , and the sets $\{C_0\}$ and $\{C_1\}$ are definable by $\Sigma _n$ -formulas with parameters in $A\cup \{\kappa \}$ . In the other direction, assume that there are closed unbounded subsets $C_0$ and $C_1$ of $\kappa $ with $C_0\cap C_1$ bounded in $\kappa $ and the property that the sets $\{C_0\}$ and $\{C_1\}$ are both definable by $\Sigma _n$ -formulas with parameters in $A\cup \{\kappa \}$ . We define $\mu =\max (C_0\cap C_1)<\kappa $ and let ${c}:{\omega }\longrightarrow {\kappa }$ denote the unique function with $c(0)=\mu $ and

$$ \begin{align*}c(2k+2-i) ~ = ~ \min(C_i\setminus(c(2k+1-i)+1))\end{align*} $$

for all $k<\omega $ and $i<2$ . The $\Sigma _n$ -Recursion Theorem then implies that c is definable by a $\Sigma _n$ -formula with parameters in $A\cup \{\kappa \}$ . Moreover, we know that c is cofinal in $\kappa $ because otherwise, $\mu <\sup _{k<\omega }c(k)\in C_0\cap C_1$ .

Proof Let C be a closed unbounded subset of $\kappa $ with the property that the set $\{C\}$ is definable by a $\Sigma _n$ -formula with parameters in $\mathrm {H}_{\mu }\cup \{\kappa ,\gamma \}$ . Assume, toward a contradiction, that $\mu $ is not an element of C. Set $\nu =\min (C\setminus \mu )>\mu $ . Then, $C\cap \mu \neq \emptyset $ because otherwise, $\nu =\min (C)$ , and this would imply that the set $\{\nu \}$ is definable by a $\Sigma _n$ -formula with parameters in $\mathrm {H}_{\mu }\cup \{\kappa ,\gamma \}$ . This allows us to define $\rho =\max (C\cap \mu )<\mu $ . But then, $\nu =\min (C\setminus (\rho +1))$ , and hence, $\{\nu \}$ is definable by a $\Sigma _n$ -formula with parameters in $\mathrm {H}_{\mu }\cup \{\kappa ,\gamma \}$ , a contradiction.

Proof Assume, toward a contradiction, that there is a $\Sigma _1$ -formula $\varphi (v_0,v_1,v_2)$ , an uncountable cardinal $\mu <\kappa $ , an ordinal $\alpha $ in the interval $[\mu ,\kappa )$ , and $z\in \mathrm {H}_{\mu }$ such that $\alpha $ is the unique ordinal $\xi $ with the property that $\varphi (\xi ,\kappa ,z)$ holds. Pick a transitive set M with $\mathrm {H}_{\kappa }\cup \{\kappa \}\subseteq M\prec _{\Sigma _1}\mathrm {H}_{\kappa ^+}$ , a transitive set N with $M\cup {}^{{<}\kappa }N\subseteq N$ , and a weakly amenable N-ultrafilter F on $\kappa $ with the property that $\langle N,\in ,F\rangle $ is iterable. Now, pick an elementary submodel $\langle X,\in ,F_0\rangle $ of $\langle N,\in ,F\rangle $ of cardinality less than $\mu $ with $\mathrm {{tc}}({\{z\}})\cup \{\kappa ,\alpha \}\subseteq X$ , and let ${\pi }:{X}\longrightarrow {N_0}$ denote the corresponding transitive collapse. Then, $\pi (z)=z$ and $\pi (\alpha )<\pi (\kappa )<\mu \leq \alpha $ . Moreover, if we set $U=\pi [F_0]$ , then U is a weakly amenable $N_0$ -ultrafilter on $\pi (\kappa )$ and [Reference Kanamori10, Theorem 19.15] implies that $\langle N_0,\in ,U\rangle $ is iterable. This yields a transitive set $N_1$ and an elementary embedding ${j}:{N_0}\longrightarrow {N_1}$ with $j(\pi (\kappa ))=\kappa $ and $j\restriction \mathrm {H}_{\pi (\kappa )}^{N_0}=\mathrm {{id}}_{\mathrm {{H}_{\pi (\kappa )}^{N_0}}}$ . Since our setup ensures that $\varphi (\alpha ,\kappa ,z)$ holds in N, we now know that $\varphi (\pi (\alpha ),\kappa ,z)$ holds in $N_1$ . By $\Sigma _1$ -upwards absoluteness, this shows that $\varphi (\pi (\alpha ),\kappa ,z)$ holds in $\mathrm {{V}}$ , contradicting the uniqueness of $\alpha $ .

Proof Fix a $\Sigma _1$ -formula $\varphi (v_0,v_1,v_2)$ and an element z of $\mathrm {H}_{\kappa }$ with the property that the set $\{E\}$ is definable by the formula $\varphi (v_0,v_1,v_2)$ and the parameters $\kappa $ and z. By our assumptions on $\kappa $ , there exists a transitive set M with $\mathrm {H}_{\kappa }\cup \{\kappa \}\subseteq M\prec _{\Sigma _1}\mathrm {H}_{\kappa ^+}$ , a transitive set N with $M\cup {}^{{<}\kappa }N\subseteq N$ , and a normal, weakly amenable N-ultrafilter F on $\kappa $ with the property that $\langle N,\in ,F\rangle $ is iterable. This setup ensures that $E\in M$ and $\varphi (E,\kappa ,z)$ holds in N. Moreover, the fact that E contains a closed unbounded subset of $\kappa $ implies that M contains such a subset of E and since F is a normal N-ultrafilter, we can conclude that E is an element of F. Pick an elementary submodel $\langle X,\in ,F_0\rangle $ of $\langle N,\in ,F\rangle $ of cardinality less than $\kappa $ with $\mathrm {{tc}}({\{z\}})\cup \{\kappa \}\subseteq X$ , and let ${\pi }:{X}\longrightarrow {N_0}$ denote the corresponding transitive collapse. Then, $\pi (z)=z$ , and if we set $U=\pi [F_0]$ , then U is a weakly amenable $N_0$ -ultrafilter on $\pi (\kappa )$ and [Reference Kanamori10, Theorem 19.15] ensures that $\langle N_0,\in ,U\rangle $ is iterable. Let

$$ \begin{align*}\langle\langle{N_\alpha}~\vert~{\alpha\in \mathrm{Ord}}\rangle,\langle{{j_{\alpha,\beta}}:{M_\alpha}\longrightarrow{M_\beta}}~\vert~{\alpha\leq\beta\in \mathrm{Ord}}\rangle\rangle\end{align*} $$

denote the linear iteration of $\langle N_0,\in ,U\rangle $ . Then, $(j_{0,\kappa }\circ \pi )(\kappa )=\kappa $ , $(j_{0,\kappa }\circ \pi )(z)=z$ , $\Sigma _1$ -upwards absoluteness implies that $\varphi ((j_{0,\kappa }\circ \pi )(E),\kappa ,z)$ holds in V, and hence, we know that $(j_{0,\kappa }\circ \pi )(E)=E$ . Moreover, the fact that $E\in F$ ensures that $\pi (E)\in U$ and therefore $(j_{0,\alpha }\circ \pi )(\kappa )\in (j_{0,\alpha +1}\circ \pi )(E)\subseteq E$ for all $\alpha <\kappa $ . This shows that the closed unbounded subset $C=\{{(j_{0,\alpha }\circ \pi )(\kappa )}~\vert ~{\alpha <\kappa }\}$ of $\kappa $ is a subset of E. Finally, the set $\{C\}$ is definable by a $\Sigma _1$ -formula with parameters $\kappa $ , $N_0$ , and U.

Proof of Theorem 4.1

The last two lemmata prove the first two assertions of the theorem. To prove the third assertion, let $\kappa $ be a stably measurable cardinal, let S be $\boldsymbol {\Sigma }_1$ -stationary in $\kappa $ , and let ${r}:{\kappa }\longrightarrow {\kappa }$ be a regressive function that is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . Our earlier observations now show that there is a unique ordinal $\alpha <\kappa $ with the property that the set $r^{{-}1}\{\alpha \}$ contains a closed unbounded subset of $\kappa $ . Since the set $\{r^{{-}1}\{\alpha \}\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ , Lemma 4.3 yields a closed unbounded subset C of $\kappa $ with $C\subseteq r^{{-}1}\{\alpha \}$ and the property that the set $\{C\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . Then, r is constant on $C\cap S$ , and since $\kappa $ is an uncountable regular cardinal, we know that this set is $\boldsymbol {\Sigma }_1$ -stationary in $\kappa $ .

Proof First, assume that $\kappa $ is not stably measurable. Then, we can apply [Reference Welch19, Theorem 2.6] to find a bistationary subset E of $\kappa $ with the property that the set $\{E\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . Lemma 3.7 then shows that the set $\{\mathrm {H}_{\kappa }\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ .

Now, assume that $\kappa $ is a limit cardinal and the set $\{\mathrm {H}_{\kappa }\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . Let C denote the set of all limit cardinals smaller than $\kappa $ . Our assumption then implies that the set $\{C\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . If C is unbounded in $\kappa $ , then the set of all successor cardinals smaller than $\kappa $ is an unbounded subset of $\kappa $ that consists of cardinals and is not $\boldsymbol {\Sigma }_1$ -stationary in $\kappa $ . In the other case, if C is bounded in $\kappa $ , then ${\mathrm {{cof}}(\kappa )}=\omega $ , Lemma 3.3 yields a cofinal function ${c}:{\omega }\longrightarrow {\kappa }$ that is definable by a $\Sigma _1$ -formula with parameter $\kappa $ , and the set $D=\{{c(n)+1}~\vert ~{n<\omega }\}$ is closed unbounded in $\kappa $ that contains no cardinals and has the property that the set $\{D\}$ is definable by a $\Sigma _1$ -formula with parameter $\kappa $ .

Together with Theorem 4.1, these computations establish all equivalences claimed in the theorem.

Proof Assume, toward a contradiction, that there is no inner model with a stably measurable cardinal and $\kappa $ is a limit cardinal with the property that every unbounded subset of $\kappa $ that consists of cardinals is $\boldsymbol {\Sigma }_1$ -stationary in $\kappa $ . Then, Corollary 3.5 shows that $\kappa $ is regular. Since $\kappa $ is not stably measurable in $\mathrm {{K}}^{DJ}$ , Theorem 4.4 now shows that in $\mathrm {K}^{DJ}$ , the set $\{\mathrm {H}_{\kappa }\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . Let C denote the set of all ordinals less than $\kappa $ that are limit cardinals in $\mathrm {{K}}^{DJ}$ . Then, C is a closed unbounded subset of $\kappa $ , and since the class $\mathrm {{K}}^{DJ}$ is definable by a $\Sigma _1$ -formula with parameter $\kappa $ , the set $\{C\}$ is definable by a $\Sigma _1$ -formula with parameters in $\mathrm {H}_{\kappa }\cup \{\kappa \}$ . Now, let E denote the set of all successor cardinals of singular cardinals less than $\kappa $ . The results of [Reference Dodd and Jensen4] then show that the elements of E are successor cardinals of singular cardinals in $\mathrm {{K}}^{DJ}$ , and hence, $C\cap E=\emptyset $ . This shows that E is an unbounded subset of $\kappa $ that consists of cardinals and is not $\boldsymbol {\Sigma }_1$ -stationary. In combination with Theorem 4.1, these arguments prove the desired equivalence.

Proof Fix an ordinal $\alpha _1<\kappa $ with the property that $j^*_{0,\alpha }\restriction \mathrm{Ord}=j^1_{0,\alpha }\restriction \mathrm{Ord}$ holds for all $\alpha <\alpha _1$ . Set $M^0=M^0_{\alpha _0}$ , $M^1=M^1_{\alpha _1}$ , $M^*=M^*_{\alpha _1}$ , $j^0=j^0_{0,\alpha _0}$ , $j^1=j^1_{0,\alpha _1}$ , $j^*=j^*_{0,\alpha _1}$ , and $U_*=j^0(U_1)$ . Then, $j^0(\kappa _1)=\kappa _1$ . In addition, if $i<2$ , then we define $\kappa ^i_\alpha =j^i_{0,\alpha }(\kappa _i)$ for all $\alpha \in \mathrm{Ord}$ and $C^i=\{{\kappa ^i_\alpha }~\vert ~{\alpha <\alpha _i}\}$ . An application of [Reference Kanamori10, Lemma 19.6] then shows that every element of $M^*$ is of the form $j^0(f)(c)$ with $n<\omega $ , ${f}:{[\kappa _0]^n}\longrightarrow {V}$ and $c\in [C^0]^n$ . In addition, given $i<2$ and $n<\omega $ , we define

$$ \begin{align*}U_i^n ~ = ~ \{{X\subseteq [\kappa_i]^n}~\vert~{\exists E\in U_i ~ [E]^n\subseteq X}\}.\end{align*} $$

The results of [Reference Kanamori10, Chapter 19] then show that $U_i^n$ is a ${<}\kappa _i$ -complete ultrafilter on $[\kappa _i]^n$ . Set $U_*^n=j^0(U_1^n)$ for all $n<\omega $ . Our assumption now implies that $j^*_{0,\alpha }(\kappa _1)=\kappa ^1_\alpha $ holds for all $\alpha <\alpha _1$ . In particular, we know that every ordinal is of the form $j^*(g)(d)$ with $n<\omega $ , ${g}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ in $M_0$ and $d\in [C^1]^n$ .

Proof of the Claim

Pick $n<\omega $ , ${f}:{[\kappa _0]^n}\longrightarrow {V}$ and $c\in [C^0]^n$ with $B_*=j^0(f)(c)$ . By [Reference Kanamori10, Lemma 19.9], we then have

$$ \begin{align*}D ~ = ~ \{{a\in[\kappa_0]^n}~\vert~{f(a)\in U_1^n}\} ~ \in ~ U_0^n.\end{align*} $$

Since $U_0^n$ has cardinality less than $\kappa _1$ , the ${<}\kappa _1$ -completeness of $U_1^n$ implies that $E=\bigcap \{{f(a)}~\vert ~{a\in D}\}$ is an element of $U_1^n$ . Pick $b\in B\cap E$ . Then,

$$ \begin{align*}D ~ \subseteq ~ \{{a\in[\kappa_0]^n}~\vert~{b\in f(a)}\} ~ \in ~ U_0^n,\end{align*} $$

and [Reference Kanamori10, Lemma 19.9] shows that $j^0(b)\in j^0[B]\cap j^0(f)(c)=j^0[B]\cap B_*\neq \emptyset $ .

Claim If $n<\omega $ and ${g_0,g_1}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ are functions in $M^0$ with

$$ \begin{align*}\{{b\in[\kappa_1]^n}~\vert~{g_0(b)=g_1(b)}\}\in U_*^n,\end{align*} $$

then

$$ \begin{align*}\{{b\in[\kappa_1]^n}~\vert~{g_0(j^0(b))=g_1(j^0(b))}\}\in U_1^n.\end{align*} $$

The same proof also yields the following implication:

Claim If $n<\omega $ and ${g_0,g_1}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ are functions in $M^0$ such that the set $\{{b\in [\kappa _1]^n}~\vert ~{g_0(b)<g_1(b)}\}$ is an element of $U_*^n$ , then the set

$$ \begin{align*}\{{b\in[\kappa_1]^n}~\vert~{g_0(j^0(b))<g_1(j^0(b))}\}\end{align*} $$

is an element of $U_1^n$ .

Claim For every function ${f}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ in V, there is a function ${g}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ in $M^0$ with

$$ \begin{align*}\{{b\in[\kappa_1]^n}~\vert~{f(b)=g(j^0(b))}\}\in U_1^n.\end{align*} $$

Proof of the Claim

Given $b\in [\kappa _1]^n$ , pick $m_b<\omega $ , ${f_b}:{[\kappa _0]^{m_b}}\longrightarrow {\mathrm{Ord}}$ in V and $c_b\in [C^0]^{m_b}$ with $f(b)=j^0(f_b)(c_b)$ . Then, there is $m<\omega $ and $c\in [C^0]^m$ with

$$ \begin{align*}B ~ = ~ \{{b\in[\kappa_1]^n}~\vert~{m_b=m, ~ c_b=c}\} ~ \in ~ U_1^n.\end{align*} $$

Define

$$ \begin{align*}{G}:{[\kappa_0]^m\times B}\longrightarrow{\mathrm{Ord}};~{(a,b)}\longmapsto{f_b(a)}\end{align*} $$

and

$$ \begin{align*}{g}:{[\kappa_1]^n}\longrightarrow{\mathrm{Ord}};~{b}\longmapsto{j^0(G)(c,b)}.\end{align*} $$

Then, g is an element of $M^0$ , and

$$ \begin{align*}g(j^0(b)) ~ = ~ j^0(G)(c,j^0(b)) ~ = ~ j^0(f_b)(c) ~ = ~ f(b)\end{align*} $$

holds for all $b\in B\in U_1^n$ .

Claim If ${g}:{[\kappa _1]^1}\longrightarrow {\mathrm{Ord}}$ is an element of $M^0$ , then

$$ \begin{align*}j^1(g\circ j^0)(\{\kappa_1\}) ~ = ~ j^1(g)(\{\kappa_1\}).\end{align*} $$

By our second claim and [Reference Kanamori10, Lemma 19.9], there exists a well-defined class function ${F}:{\mathrm{Ord}}\longrightarrow {\mathrm{Ord}}$ with the property that

$$ \begin{align*}F(j^*(g)(c)) ~ = ~ j^1(g\circ j^0)(c)\end{align*} $$

holds for all $n<\omega $ , ${g}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ in $M^0$ and $c\in [C^1]^n$ . Moreover, our third claim shows that F is order-preserving. Finally, since every ordinal is of the form $j^1(f)(c)$ for some $n<\omega $ , ${f}:{[\kappa _1]^n}\longrightarrow {\mathrm{Ord}}$ in V and $c\in [C^1]^n$ , our fourth claim shows that F is also surjective. In combination, this shows that F is the identity on $\mathrm{Ord}$ . Now, fix an ordinal $\gamma $ , and let $g_\gamma $ denote the constant function on $[\kappa _1]^1$ with value $\gamma $ . Then, we have

$$ \begin{align*} \begin{aligned} j^*(\gamma) ~ & = ~ j^*(g_\gamma)(\{\kappa_1\}) ~ = ~ F(j^*(g_\gamma)(\{\kappa_1\})) \\ & = ~ j^1(g_\gamma\circ j^0)(\{\kappa_1\}) ~ = ~ j^1(g_\gamma)(\{\kappa_1\}) ~ = ~ j^1(\gamma), \end{aligned} \end{align*} $$

and this proves the statement of the lemma.

Proof Assume, toward a contradiction, that there is an ordinal $\gamma $ such that for unboundedly many $\xi <\lambda $ , there is an $\alpha <\kappa $ with $j^\xi _{0,\alpha }(\gamma )>\gamma $ . Let $\gamma $ be minimal with this property. Pick $\zeta <\lambda $ and $\alpha _0<\kappa $ with $j^\zeta _{0,\alpha _0}(\gamma )>\gamma $ . Set $j=j^\zeta _{0,\alpha _0}$ and $M=M^\zeta _{\alpha _0}$ . Then, $j(\lambda )=\lambda $ and $j(\kappa )=\kappa $ . Moreover, if we set $j(\vec {\kappa })=\langle {\kappa ^\prime _\xi }~\vert ~{\xi <\lambda }\rangle $ and $j(\vec {U})=\langle {U^\prime _\xi }~\vert ~{\xi <\lambda }\rangle $ , then the fact that $j(\gamma )>\gamma $ yields an ordinal $\rho <\lambda $ with the property that $i^\xi _{0,\alpha }(\gamma )=\gamma $ holds for all $\rho \leq \xi <\lambda $ and all $\alpha <\kappa $ , where

$$ \begin{align*}\langle\langle{N^\xi_\alpha}~\vert~{\alpha\in \mathrm{Ord}}\rangle,\langle{{i^\xi_{\alpha,\beta}}:{N^\xi_\alpha}\longrightarrow{N^\xi_\beta}}~\vert~{\alpha\leq\beta\in \mathrm{Ord}}\rangle\rangle\end{align*} $$

denotes the linear iteration of $\langle M,\in ,U^\prime _\xi \rangle $ . By our assumption, we are now able to pick $\rho <\xi <\lambda $ and $\alpha _1<\kappa $ such that $\kappa _\xi>\alpha _0$ and $j^\xi _{0,\alpha _1}(\gamma )>\gamma $ . In this situation, Lemma 4.7 implies that $i^{j(\xi )}_{0,\alpha _1}(\gamma )=j^\xi _{0,\alpha _1}(\gamma )>\gamma $ , and since $\rho \leq \xi \leq j(\xi )<\lambda $ , this yields a contradiction.

Proof Lemma 4.8 yields $\lambda _0<\lambda $ with the property that $j^\xi _{0,\alpha }(\kappa )=\kappa $ and $j^\xi _{0,\alpha }(\gamma )=\gamma $ holds for all $\lambda _0\leq \xi <\lambda $ and $\alpha <\kappa $ . Assume, toward a contradiction, that there is a $\Sigma _1$ -formula $\varphi (v_0,\ldots ,v_3)$ , and ordinal $\lambda _0\leq \xi <\lambda $ , an element z of $\mathrm {H}_{\kappa _\xi }$ , and an ordinal $\alpha $ in the interval $[\kappa _\xi ,\kappa )$ such that the set $\{\alpha \}$ is definable by the formula $\varphi (v_0,\ldots ,v_3)$ and the parameters $\kappa $ , $\gamma $ , and z. Pick an ordinal $\beta <\kappa $ with $j^\xi _{0,\beta }(\kappa _\xi )>\alpha $ . Then, we know that $j^\xi _{0,\beta }(\alpha )>\alpha $ , and elementarity implies that $\varphi (j^\xi _{0,\beta }(\alpha ),\kappa ,\gamma ,z)$ holds in $M^\xi _\beta $ . But then, $\Sigma _1$ -upwards absoluteness implies that $\varphi (j^\xi _{0,\beta }(\alpha ),\kappa ,\gamma ,z)$ holds in V, contradicting our assumptions.

Proof Work in the setting of Lemma 4.8. Let E be an unbounded subset of $\kappa $ that consists of cardinals, and fix a $\Sigma _1$ -formula $\varphi (v_0,v_1,v_2)$ , an ordinal $\gamma $ , and $z\in \mathrm {H}_{\kappa }$ with the property that there exists a closed unbounded subset C of $\kappa $ that is the unique set x with the property that $\varphi (x,\gamma ,z)$ holds. By combining Lemma 4.8 and Lemma 4.9 with Corollary 2.13, we can find $\xi <\lambda $ such that $\kappa _\xi $ is an element of C, $z\in \mathrm {H}_{\kappa _\xi }$ , and for all $\alpha <\kappa $ , we have $j^\xi _{0,\alpha }(\kappa )=\kappa $ and $j^\xi _{0,\alpha }(\gamma )=\gamma $ . In this situation, we can find a cardinal $\kappa _\xi <\mu \in E$ with $j^\xi _{0,\mu }(\kappa _\xi )=\mu $ . Then, $\varphi (j^\xi _{0,\mu }(C),\gamma ,z)$ holds in $M^\xi _\mu $ , and since $\Sigma _1$ -upwards absoluteness implies that this statement also holds in $\mathrm {{V}}$ , it follows that $j^\xi _{0,\mu }(C)=C$ . Moreover, since $\kappa _\xi \in C$ , we can now use elementarity to conclude that $\mu =j^\xi _{0,\mu }(\kappa _\xi )\in C\cap E\neq \emptyset $ .